Problem: $g(t) = -6t$ $f(t) = 6t^{2}-2t-7+g(t)$ $h(t) = 7+g(t)$ $ h(f(2)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(2)$ . Then we'll know what to plug into the outer function. $f(2) = 6(2^{2})+(-2)(2)-7+g(2)$ To solve for the value of $f$ , we need to solve for the value of $g(2)$ $g(2) = (-6)(2)$ $g(2) = -12$ That means $f(2) = 6(2^{2})+(-2)(2)-7-12$ $f(2) = 1$ Now we know that $f(2) = 1$ . Let's solve for $h(f(2))$ , which is $h(1)$ $h(1) = 7+g(1)$ To solve for the value of $h$ , we need to solve for the value of $g(1)$ $g(1) = (-6)(1)$ $g(1) = -6$ That means $h(1) = 7-6$ $h(1) = 1$